Answer
$
\begin{aligned}
\vec{E}_C & =-(6.29 \mathrm{~N} / \mathrm{C}) \hat{\mathrm{i}}
\end{aligned}
$
Work Step by Step
For point $\mathrm{C}$, we have
$
\begin{aligned}
\vec{E}_C & =\left[\frac{q_1}{4 \pi \varepsilon_0 r_1^2}-\frac{\left|q_2\right|}{4 \pi \varepsilon_0 r_2^2}\right] \hat{\mathrm{i}}=\frac{\left(8.99 \times 10^9\right)\left(1.00 \times 10^{-12} \mathrm{C}\right)}{\left(2.00 \times 5.00 \times 10^{-2}\right)^2} \hat{\mathrm{i}}-\frac{\left(8.99 \times 10^9\right)\left|-2.00 \times 10^{-12} \mathrm{C}\right|}{\left(5.00 \times 10^{-2}\right)^2} \hat{\mathrm{i}} \\
& =-(6.29 \mathrm{~N} / \mathrm{C}) \hat{\mathrm{i}}
\end{aligned}
$
